MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}} - 2x}}{{x - \sin \,x}}$ =
- A$4$
- B$1$
- ✓$2$
- D$\frac {1}{2}$
Using $L'$ hospital rule
$ = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}} - 2}}{{1 - \cos x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{\sin x}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}}}}{{\cos x}} = 2$
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